Solving Permutation Index Homework Problems

[EugP]

Homework Statement

I'm taking a Magnetic Fields class, and the professor taught us doing cross and dot products using the permutation index. But I don't quite understand how it works completely.

I have these problems:

Given:

\vec A=\hat x + 2\hat y - 3\hat z
\vec B=3\hat x - 4\hat y
\vec C=3\hat y - 4\hat z

Find:

1) \vec A \times \vec C

2) \hat x \times \vec B

Homework Equations

The Attempt at a Solution

1) Using what I know about the permitivity constant:

(\vec A \times \vec C)=

\varepsilon_{xyz}\vec A_y \vec C_z=

But I don't know where to go from here. All I know is that \varepsilon_{xyz} = 1because indices are a cyclic permutation, but I don't know what to do next.<br /> <br /> 2) For this one I don't even know where to begin.<br /> <br /> Please someone help, any help at all would be great.

 

[Avodyne]

You can do these with rules for the cross products of the unit vectors:

\hat x\times\hat x = 0\quad\quad\hat x\times\hat y = \hat z\quad\quad\hat x\times\hat z = -\hat y

\hat y\times\hat x = -\hat z\quad\quad\hat y\times\hat y = 0\quad\quad\hat y\times\hat z = \hat x

\hat z\times\hat x = \hat y\quad\quad\hat z\times\hat y = -\hat x\quad\quad\hat z\times\hat z = 0

 

[Nevetsman]

Hi!

I'm new to the forum so I'm still getting used to the formatting techniques... instead of using the fancy mathematical symbols, I'm going to use easier assignments. In this case, E will be my permutation symbol.

Remember that the Einstein notation for vectors is incredibly useful, but it operates with an implied summation. E(xyz) is summed as x goes from 0 to 3, y goes from 0 to 3, and z goes from 0 to 3. These represent nothing more than component indexing numbers, with 1=i'th component, 2=j'th component, and 3=k'th component.

Your representation of the cross product is accurate, but don't include the vector line above each letter. E(xyz)A(y)C(z)=AxC. Using the summation, let's expand this, we get...

=E(123)A(2)C(3)1+E(132)A(3)C(2)1+E(213)A(1)C(3)2+E(231)A(3)C(1)2+
E(312)A(1)C(3)3+E(321)A(2)C(1)3

Then, for all cyclic arrangements of E -> E(123)=E(312)=E(231)=1
For all non-clyclic arrangements of E -> E(132)=E(213)=E(321)=-1

Now simply plug in the corresponding values for each component index.

Part B in your problem is solved in almost the exact same way, just remember that the x-unit vector does not have a y or z component, so all those components will be zero.

Hope this helps!

Steve

 

[HallsofIvy]

Homework Statement

I'm taking a Magnetic Fields class, and the professor taught us doing cross and dot products using the permutation index. But I don't quite understand how it works completely.

I have these problems:

Given:

\vec A=\hat x + 2\hat y - 3\hat z
\vec B=3\hat x - 4\hat y
\vec C=3\hat y - 4\hat z

Find:

1) \vec A \times \vec C

2) \hat x \times \vec B

Homework Equations

The Attempt at a Solution

1) Using what I know about the permitivity constant:

(\vec A \times \vec C)=

\varepsilon_{xyz}\vec A_y \vec C_z=

As nevetsman said, this should be
\varepsilon_{xyz} A_y C_z= where A_x and C_z are the x component of A and the z component of C, not vectors themselves.
For any set of indices \varepsilon_{ijklm} is defined to be "1 if ijklm is an even permutation of 12345, -1 if an odd permutation, 0 otherwise". There are 3!= 6 permutions of 123. 3 are even: 123, 231, and 312, 3 are odd: 132, 213, and 321.So \varepsilon_{123}= \varepsilon_{231}= \varepsilon_{321}= 1 while \varepsilon_{132}= \varepsilon_{213}= \varepsilon_{321}= -1.

Do you know how to find a 3 by 3 determinant? That's another mnenonic that might be simpler.

But I don't know where to go from here. All I know is that \varepsilon_{xyz} = 1 because indices are a cyclic permutation, but I don't know what to do next.

2) For this one I don't even know where to begin.

Please someone help, any help at all would be great.

 

[EugP]

Thanks for the replies. I accidentally left A and C as vectors, didn't mean to. And now I see what I have to do, thank you very much!